Problem: In triangle $ABC$, $A'$, $B'$, and $C'$ are on the sides $BC$, $AC$, and $AB$, respectively. Given that $AA'$, $BB'$, and $CC'$ are concurrent at the point $O$, and that $\frac{AO}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$, find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$.

Solution: Let $K_A=[BOC], K_B=[COA],$ and $K_C=[AOB].$ Due to triangles $BOC$ and $ABC$ having the same base,\[\frac{AO}{OA'}+1=\frac{AA'}{OA'}=\frac{[ABC]}{[BOC]}=\frac{K_A+K_B+K_C}{K_A}.\]Therefore, we have\[\frac{AO}{OA'}=\frac{K_B+K_C}{K_A}\]\[\frac{BO}{OB'}=\frac{K_A+K_C}{K_B}\]\[\frac{CO}{OC'}=\frac{K_A+K_B}{K_C}.\]Thus, we are given\[\frac{K_B+K_C}{K_A}+\frac{K_A+K_C}{K_B}+\frac{K_A+K_B}{K_C}=92.\]Combining and expanding gives\[\frac{K_A^2K_B+K_AK_B^2+K_A^2K_C+K_AK_C^2+K_B^2K_C+K_BK_C^2}{K_AK_BK_C}=92.\]We desire $\frac{(K_B+K_C)(K_C+K_A)(K_A+K_B)}{K_AK_BK_C}.$ Expanding this gives\[\frac{K_A^2K_B+K_AK_B^2+K_A^2K_C+K_AK_C^2+K_B^2K_C+K_BK_C^2}{K_AK_BK_C}+2=\boxed{94}.\]